Determine the convergence or divergence of the series. sum_{n=1}^inftyfrac{n}{sqrt{3n^2+3}}

Alyce Wilkinson

Alyce Wilkinson

Answered question

2020-11-08

Determine the convergence or divergence of the series.
n=1n3n2+3

Answer & Explanation

averes8

averes8

Skilled2020-11-09Added 92 answers

Consider the given series n=1n3n2+3
Divergence test series,
If limnan0 then series n=1an is diverge
And
If limnan=0 then series n=1an is is convergence or diverge
Given an=n3n2+3
Now find the limit
limnan=limnn3n2+3
=limnnn3+3n2
=limnnn3+3()2
=13
Hence, the series is divergence because limnan0

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